Unit 3 Nuclear
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Transcript of Unit 3 Nuclear
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Unit 3 Nuclear
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Go to question
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What is the result of an atom losing an alpha particle?
a. The atomic number increases and the mass number increases.
b. The atomic number decreases and the mass number increases.
c. The atomic number increases and the mass number decreases.
d. The atomic number decreases and the mass number decreases.
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a hint!!!!1st hintWhat is meant by atomic and mass number?
2nd hintAn alpha particle is a helium nucleus
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What is the result of an atom losing an alpha particle?
Correct because………….
An alpha particle is a helium nucleus. So is made up from2 protons and 2 neutrons. I.e. it has an atomic (proton)number of 2 and a mass number of 4. So the atomic number decreases and the mass number decreases.
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6g of a radioactive isotope of 60Co has a half-life of 5 yrs. how much of this isotope would be left after 20 yrs?
a. 1.2 g
b. 0.375 g
c. 0.75 g
d. 0.188 g
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a hint!!!!1st hintHow many half lives over a period of 20 years?
2nd hintStart-5years-10years-15years-20years.
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6g of a radioactive isotope of 60Co has a half-life of 5 yrs. how much of this isotope would be left after 20 yrs?
Correct because………
The half life of a radio active isotope is a measure of the time is takes for activity to be reduced by half. Over a period of 20 years, 4 half-lives would be needed for a period of 20 years.
60Co 60Co 60Co 60Co 60Co6g 3g 1.5g 0.75g 0.375g
1 2 3 4
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The process below represents
a. Nuclear fusion
b. Nuclear fission
c. Alpha emission
d. Neutron capture
392 0
1U
236+
56Ba
142
36Kr
91+
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a hint!!!!This process takes place inside stars, where Hydrogenis converted into helium.
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a hint!!!!An alpha particle has an atomic number of 2 and a mass number of 4.
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a hint!!!!Neutron capture would result in an increase in the mass number.
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The process below represents
392 0
1U
236+
56Ba
142
36Kr
91+
Correct because…………..
Nuclear fission involves the splitting of a nucleus into two nucleui.
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What is produced (X) when 238U is combined with a neutron.
n 01
U 238
+ e -1
0+X
a. 92
U 239
b. 93Pu
240
c. 92
U 240
d. 93
Pu 239
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a hint!!!!A beta particle is lost, how would this change the atomic number?
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a hint!!!!How does adding a neutron change the mass number?
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What is produced (X) when 238U is combined with a neutron.
X n
01
U 238
+ e -1
0+X93
Pu 239
Correct because……….
Adding a neutron to 238U the mass number willincrease by 1, but the atomic number (protonnumber) will not change. A beta particle is alsoemitted, this will change a neutron into a proton,so increasing the atomic number by 1. The atomic number of uranium is always 92.
X =93
Pu 239
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To a saturated solution of sodium nitrate, a furthersample of a radioactive sample of sodium nitrate was added. Compared to the original solution of sodium nitratethe radioactive intensity of the new solution will
a. have the same intensity of radiation as before.
b. have a lower intensity of radiation.
c. have a higher level of intensity of radiation than before.
d. have no detectable level of radiation.
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a hint!!!!An un-dissolved salt will be in dynamic equilibrium with thedissolved salt.
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To a saturated solution of sodium nitrate, a furthersample of a radioactive sample of sodium nitrate was added. Compared to the original solution of sodium nitrate,the radioactive intensity of the new solution will
Correct because……….
Have a higher level of intensity of radiation than before.This happens because in a saturated solution the excesssolid is in a dynamic equilibrium with the dissolved solid.So over time some of the radioactive sodium nitrate willdissolve.
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131I can be used to study diseased thyroid glands. Itshalf-life is 8 days. A patient was given 0.0053 mol of thisisotope. How many grams would be left after 16 days?
a. 0.0013 g
b. 0.087 g
c. 0.174 g
d. 0.0027 g
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a hint!!!!1st hintHow many grams is 0.0053 mol? n = mass/gfm
2nd hintHow many half lives are involved?
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131I can be used to study diseased thyroid glands. Itshalf-life is 8 days. A patient was given 0.0053 mol of thisisotope. How many grams would be left after 16 days?
Correct because………..
The mass of 131I in 0.0053g = n (number moles) = mass/gfmSo mass = n x gfm = 0.0053x131= 0.694g
Ans: c. 0.174 g
131I 131I 131I0.694g 0.347g 0.174g
8 16
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A sample of pottery was found to have 3.3 mgof 14 C. If the half-life of 14 Cis 5000 yrs how many mol of14 C did the sample of potteryhave when the pottery was made?
3.0
3.5
4.0
4.5
100 20 30
Time / yrs x 10 3
Mass of14C /mg
a. 1.1 x 10 –3 mol
b. 1.32 x 10 –2 mol
c. 5.5 x 10 –4 mol
d. 9.43 x 10 –4 mol
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1st hintHow many mg of Carbon was in the pot when it was made?
2nd hintn = mass/gfm
a hint!!!!
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A sample of pottery was found to have 3.3 mgof 14 C. If the half-life of 14 Cis 5000 yrs how many mol of14 C did the sample of potteryhave when the pottery was made?
Ans: d. 9.43 x 10 –4 mol
3.0
3.5
4.0
4.5
100 20 30
Time / yrs x 10 3
Mass of14C /mg
Correct because……..
3.3mg gives a time of 10000 years, i.e. 2 half-lives. Sothe pottery would have had 13.2 mg of 14 C when it was made. n (number of moles) = mass/formula mass13.2 mg/14 = 1.32 x 10 -2 / 14 =
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There are 287 naturally occurring isotopes of which 18 are unstable or radioactive. What would the neutron:proton ratio be for the radioactive isotope 239 Pu?
a. 1.54
b. 0.39
c. 1.00
d. 2.54
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1st hintFor lighter elements the ratio is 1:1
2nd hintFor the heavier elements the ratio of neutrons to protonsincreases.
a hint!!!!
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There are 287 naturally occurring isotopes of which 18 are unstable or radioactive. What would the neutron:proton ratio be for the radioactive isotope 239 Pu?
Correct because……..
1.54 (indicates 1.54 x more neutrons than protons)The neutron:proton ratio is an important factor in deciding whether or not a particular nuclide undergoesradioactive decay. The stable, lighter elements havea ratio of near to one. A greater number of neutrons toprotons will result in a stable nuclei.